Can anyone solve the differential equation of x(1+y^2)^1/2dx=y(1+x^2)^1/2dy?

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• rotchm
Lv 7
1 month ago

Hint: From highschool algebra, do you see that this is separable?

Separate it first.  What do you get? Answer that & we will proceed.

Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them. Too late!

• 1 month ago

x√( 1 + y² ) dx = y√( 1 + x² ) dy

x / √( 1 + x² ) dx = y / √( 1 + y² ) dy

∫x / √( 1 + x² ) dx = ∫ y / √( 1 + y² ) dy

LHS:

let u = ( 1 + x² )  such that

du/dx = 2x

dx = du/(2x)

∴

∫ x / √( 1 + x² ) dx

= ∫ ( x /√u )  *  du/(2x)

= ½ ∫ u^(-½) du

= √u

= √( 1 + x² ) + constant

RHS:

∴

∫ y / √( 1 + y² ) dy

= √( 1 + y² ) + constant

THUS:

√( 1 + y² ) = √( 1 + x² ) + C

1 + y²  = [ √( 1 + x² ) + C ]²

y² =  [ √( 1 + x² ) + C ]² - 1

y² = C² + x² + 2C√( x² + 1 )

y = ±√[ C² + x² + 2C√( x² + 1 ) ]

• 1 month ago

• 1 month ago

Looks separable to me

x * dx / (1 + x^2)^(1/2) = y * dy / (1 + y^2)^(1/2)

u = (1 + x^2)^(1/2)

u^2 = 1 + x^2

2u * du = 2x * dx

u * du = x * dx

u * du / u =>

du

v = (1 + y^2)^(1/2)

v^2 = 1 + y^2

2v * dv = 2y * dy

v * dv = y * dy

y * dy / (1 + y^2)^(1/2)

v * dv / v =>

dv

Now we have:

du = dv

Integrate

u + C = v

(1 + x^2)^(1/2) + C = (1 + y^2)^(1/2)

You can expand from there if you want.